欢迎来到天天文库
浏览记录
ID:48005524
大小:457.99 KB
页数:7页
时间:2020-01-12
《厦门市2016-2017学年度第一学期高二年级质量检测数学(理科)试题参考答案及评分标准.pdf》由会员上传分享,免费在线阅读,更多相关内容在行业资料-天天文库。
1、厦门市2016-2017学年度第一学期高二年级质量检测数学(理科)试题参考答案及评分标准一、选择题:本大题共12小题,每小题5分,共60分.1~5:ADCDB6~10:CBCAC11~12:BD第12题解析:连结PF,1∵OFOFOPPFc,OFP60,1222∴FPF90,PF3c,122c2又∵PF2ac,∴2ac3c,31,2a3122bc2∴11(31)233,22aa2222x0y0xy设P(x,y),M(x,y),N(x,y)为椭圆C上的点,则1,1,00002222aba
2、b222b222b22即y(ax),y(ax),0202aa222b22b22b2222(ax)(ax0)(x0x)2yyyyyya2a2a2b000kK(233)NPNM2222222xxxxxxxxxxa00000∵k3,∴K23.NPNM2b(另解:取PN中点Q,kKkK,转化为中点弦问题,使用点差法即可)NPNMNPOQ2a二、填空题:本大题共4小题,每小题5分,共20分.22313.若mn0,则mn014.12715.516.2第16题解析:S2SBD2
3、CDABDACDAD是BAC的角平分线,由角平分线定理,得AB2AC在RtAHD中,AD2AH,则ADH30设CDx,ACy法一:由余弦定理:2221(2y)(2x)在ABD中,cos1504y2221yx33在ACD中,cos30解方程组:xBC2y62法二:设由BAD,由正弦定理:2y1在ABD中,,sin150sin(30)y1ACD中sin30sin(150)22解得:cos33sin,结合sincos1,7321解得sin,cos141
4、4高二数学(理科)试题参考答案及评分标准第1页共7页21y,6yx由正弦定理,ACD中,sin30sin33解得xBC623法三:在RtADH中,DH,由勾股定理:22321223212在RtAHC中,y(x)(),在RtAHB中,(2y)(2x)()222233解方程组:xBC62三、解答题:本大题共6小题,共70分.17.本小题考查正、余弦定理、三角形面积公式、两角和三角公式;考查计算求解能力、推理论证能力能力;考查方程思想.本小题满分10分.abc解:法一:(Ⅰ)由正弦定理:2R,sinAsi
5、nBsinC有a2RsinA,b2RsinB,c2RsinC则2cosA(2RsinAcosB2RsinBcosA)2RsinC2cosA(sinAcosBsinBcosA)sinC·····························································1分2cosAsin(AB)sinC···············································································2分sin(AB)s
6、in(C)sinC2cosAsinCsinC·························································································3分0C有sinC02cosA1解得A··········································4分3(Ⅱ)由(Ⅰ)知A,又a7322由余弦定理得:bcbc49(*)········································6分1ABC的面
7、积SbcsinA103,即bc40(**)········································7分2222由(*)(**)得,bcbc(bc)3bc49,··············································8分解得bc13,································································································9分ABC的周长为abc20.····
8、·····························································
此文档下载收益归作者所有